Question

Write the center and radius for the following circles

Answer 1

Given:

The equations of circle are

4. $\left(x+\dfrac{2}{3}\right)^2+(y+1)^2=24$

5. $(x-1)^2+(y+2)^2=28$

6 .$(x+12)^2+y^2=\dfrac{3}{64}$

To find:

The center and radius for the given circles.

Solution:

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where, $(h,k)$ is center and $r$ is the radius of the circle.

4.

The equation of the circle is

$\left(x+\dfrac{2}{3}\right)^2+(y+1)^2=24$

The standard form of this circle is

$\left(x-(-\dfrac{2}{3})\right)^2+(y-(-1))^2=(\sqrt{24})^2$

$\left(x-(-\dfrac{2}{3})\right)^2+(y-(-1))^2=(2\sqrt{6})^2$

Therefore, the center of the circle is $\left(-\dfrac{2}{3},-1\right)$ and the radius of the circle is $2\sqrt{6}$.

5.

The equation of the circle is

$(x-1)^2+(y+2)^2=28$

The standard form of this circle is

$(x-1)^2+(y-(-2))^2=(\sqrt{28})^2$

$(x-1)^2+(y-(-2))^2=(2\sqrt{7})^2$

Therefore, the center of the circle is $\left(1,-2\right)$ and the radius of the circle is $2\sqrt{7}$.

6.

The equation of the circle is

$(x+12)^2+y^2=\dfrac{3}{64}$

The standard form of this circle is

$(x-(-12))^2+(y-0)^2=(\sqrt{\dfrac{3}{64}})^2$

$(x-(-12))^2+(y-0)^2=\dfrac{\sqrt{3}}{8}$

Therefore, the center of the circle is $\left(-12,0\right)$ and the radius of the circle is $\dfrac{\sqrt{3}}{8}$.